1. **Problem:** Prove algebraically that for all positive integers $n$, the expression $n^3 + 3n^2 - 2n$ is always even. 2. **Formula and rules:** An integer is even if it is divisible by 2. We will factor the expression and show it contains a factor of 2. 3. **Factor the expression:** $$n^3 + 3n^2 - 2n = n(n^2 + 3n - 2)$$ 4. **Factor the quadratic:** $$n^2 + 3n - 2 = (n + 2)(n - 1)$$ 5. **Rewrite the expression:** $$n(n + 2)(n - 1)$$ 6. **Analyze parity:** Among three consecutive integers $n-1$, $n$, and $n+2$, at least one is even. 7. **Conclusion:** Since the product includes an even factor, the entire product is even for all positive integers $n$. **Final answer:** $n^3 + 3n^2 - 2n$ is always even for all positive integers $n$.